Question

[Le nguyen phuc]

cho M =2/2^1+3/2^2+4/2^3+...+2017/2^2016+2018/2^2017

so sanhsM với 3

Answer 1

[Le nguyen phuc]

giúp bố với bay

Answer 2

[do thi kieu oanh]

Ta có : 2M = 2 +\(\frac{3}{2}\)+\(\frac{4}{2^2}\)+...+\(\frac{2017}{2^{2015}}\)+ \(\frac{2018}{2^{2016}}\)

 2M - M = 2 + \(\frac{3}{2}\)- \(\frac{2}{2}\)+ \(\frac{4}{2^2}\)-\(\frac{3}{2^2}\)+...+\(\frac{2017}{2^{2015}}\)-\(\frac{2016}{2^{2015}}\)+ \(\frac{2018}{2^{2016}}\)-\(\frac{2017}{2^{2016}}\)-\(\frac{2018}{2^{2017}}\)

 M = 2 + \(\frac{1}{2}\)+\(\frac{1}{2^2}\)+...+\(\frac{1}{2^{2015}}\)+ \(\frac{1}{2^{2016}}\)-\(\frac{2018}{2^{2017}}\)

 Đặt N = \(\frac{1}{2}\)+\(\frac{1}{2^2}\)+...+\(\frac{1}{2^{2016}}\)

Ta có :2N = 1 + \(\frac{1}{2}\)+\(\frac{1}{2^2}\)+ .....+\(\frac{1}{2^{2015}}\)

2N - N = 1\(\frac{1}{2^{2016}}\)

Vậy N < 1 

Nên M < 2 + 1 - \(\frac{2018}{2^{2017}}\)= 3 -\(\frac{2018}{2^{2017}}\)

Vậy M < 3